16 Why special relativity? (II)

Let dx, dy, dz, and dt be the coordinate intervals associated with an infinitesimal path segment dC. Another important consequence of the transformation (2.14.1) is that the expression

(2.15.1) ds2 = dt2 + K(dx2 + dy2 + dz2)

is invariant under it: the value of ds2 does not change if dx, dy, dz, and dt are replaced by dx’, dy’, dz’, and dt’. If K=0, this simply means that the same time coordinate can be used for all inertial frames; t is the “absolute time” of Newtonian physics.

To discover the meaning of ds if K<0, we consider a clock that follows a spacetime path C. During any infinitesimal time interval, the increment of time it shows equals the increment of the time coordinate of any inertial frame relative to which it is “momentarily” (that is: during that infinitesimal time interval) at rest. Since dx=0, dy=0, and dz=0 in such a frame, ds2 equals dt2 in such a frame. If the clock travels with variable speed, the time that passes “on” it — its proper time — differs from inertial time regardless of which inertial frame is used.

Observe that if K is positive, Eq. (2.15.1) contains nothing that makes the time coordinate distinct from any space coordinate. (To make this completely obvious, choose units in which K=1.) This confirms our previous conclusion that no objective difference exists between a space axis and the time axis if K>0.

If K≤0, there exists an invariant speed. (Whatever moves with an invariant speed relative to one inertial frame, moves with exactly the same speed relative to any other inertial frame.)

If K=0, the invariant speed is infinite: what travels with infinite speed relative to one inertial frame, does so relative to every inertial frame. Differently put, if the departure of an object in one place and its arrival in another are simultaneous with respect to one inertial frame, they are so with respect to every inertial frame. Simultaneity is absolute.

If K<0, the invariant speed is finite and is given by c = 1/√(−K). Infinite or finite, the invariant speed cannot be attained by any object that can be at rest, inasmuch as this would require an infinite supply of energy. And since one has to attain the invariant speed before one can make a U–turn in a spacetime plane containing the time axis, it is the existence of an invariant speed that makes it impossible to go back in time by performing such a U–turn.

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The case against K = 0

The invariant interval of proper time ds is the “length” of the path segment dC — the function of dx, dy, dz, and dt that we were looking for. With it, the propagator for a free and stable particle (2.13.2) takes the form

What jumps out at us straightaway is that if K were zero, every spacetime path from (A,tA) to (B,tB) would contribute the same amplitude [1: −b∫Cdt] = [1: −b(tB−tA)] to the propagator (2.15.2), which would be infinite and therefore physically meaningless as a result — as well as independent of the distance between A and B! If it is to yield finite probabilities, cancellations (“destructive interference“) must occur. The complex numbers [1: −bs(C)] = [1: −b∫Cds] must not be the same for every path C from (A,tA) to (B,tB). Considered as vectors in a plane, they must not all point in the same direction. (While this does not quite rule out the option K=0 — after all, there is such a thing as non-relativistic quantum mechanics — the manner in which the non-relativistic theory is obtained from the relativistic one makes it clear that the former can only be an approximation. It is useful as long as the particle’s speed v is so small compared to the invariant speed c that all powers of v2/c2 but the first can be ignored.)

With K = −1/c2 < 0, the transformations (2.14.1) become the famous Lorentz transformations, which lie at the heart of the special theory of relativity:

It should be stressed that we have derived this transformation law without invoking the constancy of the speed of light. Instead, we found that a finite invariant speed must exist. This turns out to be the speed with which light propagates (in vacuum).

It may not have escaped your notice that the proper time interval ds determined by Eq. (2.15.1) may be positive as well as negative. As it happens, it is both. There are two types of particles: the “ordinary” particles, for which it is positive, and their so-called antiparticles, for which it is negative.[1]